JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Permutations and Combinations

The digits $2,1,2,3$ form $\frac{4!}{2!}=12$ distinct numbers. In any fixed place, $1$ appears $\frac{3!}{2!}=3$ times, $3$ appears $3$ times, and $2$ appears $12-3-3=6$ times, so the digit sum in each place is $1\cdot 3+3\cdot 3+2\cdot 6=24$. Each place value contributes equally, so the total is $24\times (1000+100+10+1)=24\times 1111=26664$.